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3 years ago

The mass of an electron is...

1 answer:

4 0

<span>The mass of an electron is not significant to the overall mass of the atom. (B)

It takes roughly 1,850 electrons to equal the mass of ONE proton or neutron,
but the most complex atom has only around 100 of them.</span>

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A current of 9 A flows through an electric device with a resistance of 43 Ω. What must be the applied voltage in this particular

blondinia [14]

The answer is D. Hope this helps you!

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3 years ago

A coin purse contains d dimes and q quarter's. There are 20 coins in the purse and the total value of the coins is $4.25. Which

ch4aika [34]

Answer:

Explanation:

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2 years ago

A proton moving at 3.90 106 m/s through a magnetic field of magnitude 1.80 T experiences a magnetic force of magnitude 7.20 10-1

VARVARA [1.3K]

Answer:

$\theta=40^0$

Explanation:

The magnitude of the magnetic force is

$F_m=evB\sin\theta$

To find the angle, we make $\sin\theta$ subject of the formula

$\implies \sin\theta=\frac{F_m}{evB}=\frac{7.20\times 10^{-13}}{1.6\times 10^{-19}\times 3.90\times 10^6\times 1.80}$

$\implies \sin\theta=0.641025641$

$\therefore \theta=\sin^{-1}=39.8683^0\\\implies \theta\approxeq 40^0$

8 0

3 years ago

A 0.320 kg ball approaches a bat horizontally with a speed of 14.0 m/s and after getting hit by the bat, the ball moves in the o

katen-ka-za [31]

Answer:

Explanation:

Step one:

<u>Given </u>

mass m= 0.32kg

intital velocity, u= 14m/s

final velocity v= 22m/s

time= 0.06s

Step two:

<u>Required</u>

Force F

the expression for the force is

F=mΔv/t

F=0.32*(22-14)/0.06

F=(0.32*8)/0.06

F=2.56/0.06

F=42.67N

The average force exerted on the bat 42.67N

4 0

2 years ago

outward from a wall just above floor level. A 1.5 kg box sliding across a frictionless floor hits the end of the spring and comp

sweet [91]

Answer:

v = 0.489 m/s

Explanation:

It is given that,

Mass of a box, m = 1.5 kg

The compression in the spring, x = 6.5 cm = 0.065 m

Let the spring constant of the spring is 85 N/m

We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2$

$v=\sqrt{\dfrac{kx^2}{m}} \\\\v=\sqrt{\dfrac{85\times (0.065)^2}{1.5}} \\\\v=0.489\ m/s$

So, the speed of the box is 0.489 m/s.

3 0

3 years ago